What is the general solution for cot^2(θ)+1= 3/4 ?

The general solution for cot^2(θ)+1= 3/4 is No Solution for θ\in\mathbb{R}

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Popular >

cot^2(θ)+1= 3/4

Solution

cot^{2} (θ) + 1 = \frac{3}{4}

Solution

No Solution for θ \in R

Solution steps

cot^{2} (θ) + 1 = \frac{3}{4}

Solve by substitution

cot^{2} (θ) + 1 = \frac{3}{4}

Let:

cot (θ) = u

u^{2} + 1 = \frac{3}{4}

u^{2} + 1 = \frac{3}{4} : u = i \frac{1}{2}, u = - i \frac{1}{2}

u^{2} + 1 = \frac{3}{4}

Move

1

to the right side

u^{2} + 1 = \frac{3}{4}

Subtract

1

from both sides

u^{2} + 1 - 1 = \frac{3}{4} - 1

Simplify

u^{2} = \frac{3}{4} - 1

u^{2} = \frac{3}{4} - 1

Simplify

\frac{3}{4} - 1 : - \frac{1}{4}

\frac{3}{4} - 1

Convert element to fraction:

1 = \frac{1 \cdot 4}{4}

= - \frac{1 \cdot 4}{4} + \frac{3}{4}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 \cdot 4 + 3}{4}

- 1 \cdot 4 + 3 = - 1

- 1 \cdot 4 + 3

Multiply the numbers:

1 \cdot 4 = 4

= - 4 + 3

Add/Subtract the numbers:

- 4 + 3 = - 1

= - 1

= \frac{- 1}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - \frac{1}{4}, u = - - \frac{1}{4}

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

- \frac{1}{4} = - 1 \frac{1}{4}

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{1}{2}

Apply rule

1 = 1

= i \frac{1}{2}

Rewrite

i \frac{1}{2}

in standard complex form:

\frac{1}{2} i

i \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2}

Multiply:

1 i = i

= \frac{i}{2}

= \frac{1}{2} i

Simplify

- - \frac{1}{4} : - i \frac{1}{2}

- - \frac{1}{4}

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

- \frac{1}{4} = - 1 \frac{1}{4}

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{1}{2}

= - i \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2} i

u = i \frac{1}{2}, u = - i \frac{1}{2}

Substitute back

u = cot (θ)

cot (θ) = i \frac{1}{2}, cot (θ) = - i \frac{1}{2}

cot (θ) = i \frac{1}{2}, cot (θ) = - i \frac{1}{2}

cot (θ) = i \frac{1}{2} :

No Solution

cot (θ) = i \frac{1}{2}

No Solution

cot (θ) = - i \frac{1}{2} :

No Solution

cot (θ) = - i \frac{1}{2}

No Solution

Combine all the solutions

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cot^2(θ)+1= 3/4 ?
The general solution for cot^2(θ)+1= 3/4 is No Solution for θ\in\mathbb{R}